Two drops of the same radius are falling thro | vedclass

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(c)If two drops of same radius r coalesce then radius of new drop is given by R
$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$=

Vedclass · Accepted Answer

$5 	imes {(4)^{1/3}}cm$ per sec

Vedclass · Answer

(c)If two drops of same radius r coalesce then radius of new drop is given by R
$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$==> ${R^3} = 2{r^3} \Rightarrow R = {2^{1/3}}r$
If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto {r^2}$
$\frac{{{v_2}}}{{{v_1}}} = {\left( {\frac{R}{r}} ight)^2} = {\left( {\frac{{{2^{1/3}}r}}{r}} ight)^2}$
==> ${v_2} = {2^{2/3}} 	imes {v_1} = {2^{2/3}} 	imes (5) = 5 	imes {(4)^{1/3}}m/s$

Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$ If the two drops coalesce, the terminal velocity would be

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